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For which real numbers $p>0$ does the series $$\sum_{n=1}^{\infty} \frac{\cos(n^p \pi)}{n^p}$$ converge?

Obviously it converges absolutely for $p>1$ but what about $0<p<1$? I have the feeling that something qualitative happens at $p=1/2$.

Help wanted, thanks.

Georgy
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  • I think the Euler's formula $e^{ix} = cos{x} + isin{x}$ might be helpful here. Considering $\sum_{n=1}^{N} \cos(n^p \pi) = Re(\sum_{n=1}^Ne^{in^p \pi})$ and trying to find out the upper bound of $\sum_{n=1}^{N} \cos(n^p \pi)$. – sundaycat Apr 26 '14 at 20:13
  • I tried that and it doesn't work. I think it's more like the denominator slows the thing down. – Georgy May 12 '14 at 05:08

2 Answers2

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We restrict our attention to the case $0 < p < 1$, since the answer is well-known for $p \geq 1$. Using the Riemann-Stieltjes integral, we can rewrite the partial sum as

$$ \sum_{k=1}^{n} \frac{\cos (\pi k^{p})}{k^{p}} = \int_{1^{-}}^{n} \frac{\cos \pi x^{p}}{x^{p}} d [x] = \int_{1^{-}}^{n^{p}} \frac{\cos \pi x}{x} d [x^{1/p}]. \tag{1} $$

Using the periodic Bernoulli polynomials $B_{m}(x)$, we have

$$ \sum_{k=1}^{n} \frac{\cos (\pi k^{p})}{k^{p}} = - \int_{1^{-}}^{n^{p}} \frac{\cos \pi x}{x} d B_{1}(x^{1/p}) + \int_{1}^{n^{p}} \frac{\cos \pi x}{x} \, d(x^{1/p}) =: I_{n} + J_{n}. $$

Step 1. Note first that

$$ J_{n} = \frac{1}{p} \int_{1}^{n^{p}} \frac{\cos \pi x}{x^{2-(1/p)}} \, dx. $$

This converges as $n \to \infty$ if and only if $2-(1/p) > 0$, or equivalently, $p > 1/2$. Indeed,

  • For $1/2 < p < 1$, the convergence follows by usual integration-by-parts trick.
  • For $0 < p \leq 1/2$, let $$ a_{k} = \lceil (2\pi k - (\pi/4))^{1/p} \rceil \quad \text{and} \quad b_{k} = \lfloor (2\pi k + (\pi/4))^{1/p} \rfloor. $$ Then we observe that $a_{k}^{p} - (2\pi k - (\pi/4)) \to 0$ and $b_{k}^{p} - (2\pi k + (\pi/4)) \to 0$. Thus for large $k$, $$ J_{b_{k}} - J_{a_{k}} = \frac{1}{p} \int_{a_{k}^{p}}^{b_{k}^{p}} \frac{\cos \pi x}{x^{2-(1/p)}} \, dx \gtrsim \int_{-\pi/4}^{\pi/4} (x+2\pi k)^{(1/p)-2} \, dx \gtrsim 1,$$ which cannot happen if $J_{n}$ converges. (That is, $J_{n}$ is not Cauchy and hence not convergent.)

Step 2. Now we look into the term $I_{n}$. Integration by parts shows that

$$ I_{n} = - \left[ \frac{\cos \pi x}{x} B_{1}(x) \right]_{1^{-}}^{n^{p}} - \int_{1}^{n^{p}} \left( \frac{\pi \sin \pi x}{x} + \frac{\cos \pi x}{x^{2}} \right) B_{1}(x^{1/p}) \, dx. $$

Here, the only term whose convergence is unclear is

$$ \int_{1}^{n^{p}} \frac{\pi \sin \pi x}{x} B_{1}(x^{1/p}) \, dx = p \pi \int_{1}^{n} \frac{\sin (\pi x^{p})}{x} B_{1}(x) \, dx. $$

Introducing the function

$$ C(x) = -\int_{x}^{\infty} \frac{B_{1}(t)}{t} \, dt = \mathcal{O}\left( \frac{1}{x} \right), $$

it follows that

$$ p \pi \int_{1}^{n} \frac{\sin (\pi x^{p})}{x} B_{1}(x) \, dx = p \pi \left[ \sin (\pi x^{p}) C(x) \right]_{1}^{n} - p^{2}\pi^{2} \int_{1}^{n} x^{p-1} \cos(\pi x^{p}) C(x) \, dx, $$

which converges absolutely as $n \to \infty$. Putting together, $I_{n}$ converges for any $0 < p < 1$ and therefore the sum (1) converges if and only if $p > 1/2$.

Remark. We have the same answer if cosine is replaced by sine.

Remaining questions.

  1. Asymptote for the infinite sum as $p \downarrow 1/2$.
  2. Same question for the sum $$ \sum_{k=1}^{\infty} \frac{\cos (a k^{p})}{k^{q}}, $$ where $p$ and $q$ are now allowed to have different values.
Sangchul Lee
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  • I am not familiar with your notation in $$\int_{1^{-}}^{n} \frac{\cos \pi x^{p}}{x^{p}} d [x] , \int_{1^{-}}^{n^{p}} \frac{\cos \pi x}{x} d [x^{1/p}]. \tag{1}$$. What is the meaning of $\int d [x]$ and $\int d [x^{1/p}]$. – Georgy May 12 '14 at 07:09
  • +1 brilliant, 初音ミク? – achille hui May 12 '14 at 07:18
  • Yes I am aware of that but what's the meaning of it as an integration variable when it's not not differentiable? Is it a sum of delta functions, is this what you mean? – Georgy May 12 '14 at 07:25
  • @sos440 This might be related. http://math.stackexchange.com/questions/215528/two-questions-about-weakly-convergent-series-related-to-sinn2-and-weyls-i – Sungjin Kim May 12 '14 at 07:28
  • @Georgy sos440 is using Riemann Stieltjes integral. The integrator need not be differentiable everywhere. – achille hui May 12 '14 at 07:34
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    @Georgy, For a continuous function $f$ and a step function $F(x)$ with jumps of size $d_{j}$ at each $x_{j}$, we have $$ \int_{a^{-}}^{b^{+}} f(x) , dF(x) = \sum {a \leq x{j} \leq b} f(x_{j}) d_{j}. $$ In particular, $d[x]$ is identical to the Dirac comb. But a good aspect of this notation is that we can perform integration by part. You can also see the link which achille hui is referring to. – Sangchul Lee May 12 '14 at 07:36
  • @achillehui, Yes, exactly. :) – Sangchul Lee May 12 '14 at 07:37
  • @sos440 For the first one in remaining questions, it looks like you will have an expression involving $\cos\pi n^p$, $\sin\pi n^p$, and a converging integral, after repeated use of IBP. – Sungjin Kim May 12 '14 at 07:38
  • @i707107, That makes sense. Since only the term $J_{\infty}$ contributes to the divergence as $p \downarrow 1/2$, we may perform IBP to get an asymptotic expansion for the series. – Sangchul Lee May 12 '14 at 07:50
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    In my course we proved that the series $\sum sin(\frac{\pi\sqrt{n}}{n^\alpha})$ converges if and only if $\alpha>1/2$. A long proof, so this answer is very awesome. –  May 12 '14 at 10:50
  • @Edwin, That also looks nice! I guess that maybe we can first estimate the growth rate of the partial sum $\sum_{k \leq n} \cos(\pi k^{p})$ and then apply some well-known tests such as Abel test. – Sangchul Lee May 12 '14 at 16:15
  • @sos440 Yes, we can apply integral test,using integration by parts before $\int_{n}^{n+1}f(t)dt-f(n)= \int_{n}^{n+1}(n+1-t)f'(t)dt$ for the case $\alpha\in(\frac{1}{2},1]$, for the case $\alpha\leq 1/2$ using cauchy criterion. In fact we can prove that if $f\in C^1$ such that $f$ and $f'$ are in $L^1([0,+\infty))$ then $\sum f(n)$ converges :) –  May 12 '14 at 16:41
  • @sos440, Great proof!! I didn't know there was such huge machinery behind approximating partial sums and applicable to so many cases. – Georgy May 12 '14 at 20:39
  • @sos440, I don't quite follow your reasoning about $a_k^p -(2 \pi k- \pi/4) \to 0$. In fact when e.g $p=1/4$ it looks like this limit is $- \pi/2$ – Georgy May 16 '14 at 15:46
  • @Georgy, let $x_{k} = (2\pi k - \pi/4)^{1/p}$ so that $a_{k} = \lceil x_{k} \rceil$. Then $$ a_{k}^{p} = (x_{k} + \mathcal{O}(1))^{p} = x^{p}{ 1 + \mathcal{O}(1/x_{k}) }^{p} = x^{p}{ 1 + \mathcal{O}(1/x_{k}) } = x_{k}^{p} + o(1). $$ – Sangchul Lee May 17 '14 at 19:08
  • I have a hunch that this $1/2$ is related to the critical line $z=1/2+it$ of the Reimann zeta function, but it looks unlikely for me to demonstrate that link. – Georgy Jun 11 '14 at 08:30
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I. The series is absolutely convergent if and only if $p>1$. This is trivial for $p>1$. If $p\le1$ then the half of the terms we have $|\cos(..)|\ge\frac1{\sqrt2}$ and $\sum\frac1{n^p}$ diverges.

II. We prove that the series diverges for $0<p\le\frac12$.

Estimate a partial sum of consecutive terms where $\cos(n^p\pi)\ge\frac12$. Take a large integer $k$ and consider those indices $n$ that satisfy $2k-\frac13\le n^p\le 2k+\frac13$; so let $$ S_k = \sum_{2k-\frac13\le n^p \le 2k+\frac13} \frac{\cos(n^p\pi)}{n^p}. $$ The number of terms in $S_k$ is roughly $$ (2k+\frac13)^{1/p}-(2k-\frac13)^{1/p} \approx C_1\cdot k^{\frac1p-1} $$ with some positive constant $C_1$; the magnitude order of the terms is $k^{-1}$. Hence, $$ S_k > C_2 k^{\frac1p-2}. $$ The last exponent, $\frac1p-2$ is nonnegative. So $S_k$ has a positive lower bound, so $S_k\not\to0$. Therefore, the series diverges.

III. Now assume $\frac12<p<1$. Replace the series by the improper integral. It is easy to verify that $$ \left|\frac{\cos(n^p\pi)}{n^p} - \int_{n-\frac12}^{n+\frac12}\frac{\cos(x^p\pi)}{x^p}\,dx \right| \le C_3 \max_{n\le x\le n+1} \left|\bigg(\frac{\cos(x^p\pi)}{x^p}\bigg)''\right| < \frac{C_4}{n^{2-p}}. $$ Since $\sum\limits\frac1{n^{2-p}}$ is convergent, we can conclude that $\sum\limits_{n=1}^\infty\frac{\cos(n^p\pi)}{n^p}$ is convergent if and only if $\int\limits_{1}^\infty\frac{\cos(x^p\pi)}{x^p}dx$ is convergent.

Integrating by parts, $$ \int_1^\infty \frac{\cos(x^p\pi)}{x^p}dx = \frac1{p\pi}\int_1^\infty (\sin(x^p\pi))' \frac1{x^{2p-1}}dx = $$ $$ \frac1{p\pi}\left[\frac{\sin(x^p\pi)}{x^{2p-1}}\right]_1^\infty +\frac{2p-1}{p\pi} \int_1^\infty \frac{\sin(x^p\pi)}{x^{2p}} dx. $$ Due to $2p>1$ the last integral is absolutely convergent.

G.Kós
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